数学题的解答，急啊，谢谢了

y=sinx+cosx+2sinxcosx+2=(sinx+cosx)²+(sinx+cosx)+1
令t=sinx+cosx,-√2<=t<=√2
y=t^2+t+1=(t+1/2)^2+3/4>=3/4
当t=-1/2时等号成立，取最小值y=3/4
由对称轴及单调性知t=√2时取最大值y=3+√2

最什么值？

解：y=sinx+cosx+2sinxcosx+2=(sinx+cosx)²+(sinx+cosx)+1
=[(sinx+cosx)+(1/2)]²+3/4=[√2sin(x+π/4)+1/2]²+3/4
∵x∈R，∴当x+π/4=π/2+2kπ，即x=π/4+2kπ时，有y(max)=3+√2
当x+π/4=3π/2+2kπ,即x=5π/4+2kπ时，有y(min)=3-√2

y=sinx+cosx+2sinxcosx+2
=(sinx+cosx)+[2sinxcosx+(sinx)^2+(cosx)^2]+1
=(sinx+cosx)+(sinx+cosx)^2+1
=[(√2)sin(x+π/4)+1/2]^2+3/4

当x=2kπ+π/4时,
y取最大值=3+√2
当(√2)sin(x+π/4)=-1/2时,即x=2kπ-arc sin[(√2)/4]-π/4
y取最小值=3/4